Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
2
$begingroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
asked 5 hours ago
Arvin DingArvin Ding
134
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Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
2
$begingroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
asked 5 hours ago
Arvin DingArvin Ding
134
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
5 hours ago
2
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
5 hours ago
1
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
5 hours ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
5 hours ago
add a comment |
2
2
2
$begingroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
asked 5 hours ago
Arvin DingArvin Ding
134
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
discrete-mathematics intuition pigeonhole-principle
asked 5 hours ago
Arvin DingArvin Ding
134
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Arvin DingArvin Ding
134
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Arvin DingArvin Ding
134
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Arvin DingArvin Ding
134
asked 5 hours ago
Arvin DingArvin Ding
134
134
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
5 hours ago
2
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
5 hours ago
1
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
5 hours ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
5 hours ago
add a comment |
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
5 hours ago
2
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
5 hours ago
1
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
5 hours ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
5 hours ago
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
5 hours ago
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
5 hours ago
2
2
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
5 hours ago
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
5 hours ago
1
1
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
5 hours ago
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
5 hours ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
5 hours ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
7
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
1
$begingroup$
Why did you start with 2, 4, and 6 rather than 1, 2, and 3?
$endgroup$
– Arcanist Lupus
2 hours ago
$begingroup$
@ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above.
$endgroup$
– JMoravitz
2 hours ago
1
$begingroup$
@ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
{1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3cdot 2^𝑘$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc.
$endgroup$
– smci
1 hour ago
$begingroup$
Vaguely related: Highly composite number
$endgroup$
– smci
1 hour ago
add a comment |
8
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
answered 4 hours ago
saulspatzsaulspatz
17k31434
$endgroup$
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
1
$begingroup$
Why did you start with 2, 4, and 6 rather than 1, 2, and 3?
$endgroup$
– Arcanist Lupus
2 hours ago
$begingroup$
@ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above.
$endgroup$
– JMoravitz
2 hours ago
1
$begingroup$
@ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
{1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3cdot 2^𝑘$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc.
$endgroup$
– smci
1 hour ago
$begingroup$
Vaguely related: Highly composite number
$endgroup$
– smci
1 hour ago
add a comment |
7
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
1
$begingroup$
Why did you start with 2, 4, and 6 rather than 1, 2, and 3?
$endgroup$
– Arcanist Lupus
2 hours ago
$begingroup$
@ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above.
$endgroup$
– JMoravitz
2 hours ago
1
$begingroup$
@ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
{1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3cdot 2^𝑘$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc.
$endgroup$
– smci
1 hour ago
$begingroup$
Vaguely related: Highly composite number
$endgroup$
– smci
1 hour ago
add a comment |
7
7
7
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
answered 4 hours ago
JMoravitzJMoravitz
48.2k33886
48.2k33886
1
$begingroup$
Why did you start with 2, 4, and 6 rather than 1, 2, and 3?
$endgroup$
– Arcanist Lupus
2 hours ago
$begingroup$
@ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above.
$endgroup$
– JMoravitz
2 hours ago
1
$begingroup$
@ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
{1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3cdot 2^𝑘$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc.
$endgroup$
– smci
1 hour ago
$begingroup$
Vaguely related: Highly composite number
$endgroup$
– smci
1 hour ago
add a comment |
1
$begingroup$
Why did you start with 2, 4, and 6 rather than 1, 2, and 3?
$endgroup$
– Arcanist Lupus
2 hours ago
$begingroup$
@ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above.
$endgroup$
– JMoravitz
2 hours ago
1
$begingroup$
@ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
{1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3cdot 2^𝑘$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc.
$endgroup$
– smci
1 hour ago
$begingroup$
Vaguely related: Highly composite number
$endgroup$
– smci
1 hour ago
1
1
$begingroup$
Why did you start with 2, 4, and 6 rather than 1, 2, and 3?
$endgroup$
– Arcanist Lupus
2 hours ago
$begingroup$
Why did you start with 2, 4, and 6 rather than 1, 2, and 3?
$endgroup$
– Arcanist Lupus
2 hours ago
$begingroup$
@ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
@ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above.
$endgroup$
– JMoravitz
2 hours ago
1
1
$begingroup$
@ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
@ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration.
$endgroup$
– JMoravitz
2 hours ago
$begingroup$
{1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3cdot 2^𝑘$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc.
$endgroup$
– smci
1 hour ago
$begingroup$
{1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3cdot 2^𝑘$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc.
$endgroup$
– smci
1 hour ago
$begingroup$
Vaguely related: Highly composite number
$endgroup$
– smci
1 hour ago
$begingroup$
Vaguely related: Highly composite number
$endgroup$
– smci
1 hour ago
add a comment |
8
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
answered 4 hours ago
saulspatzsaulspatz
17k31434
$endgroup$
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
4 hours ago
add a comment |
8
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
answered 4 hours ago
saulspatzsaulspatz
17k31434
$endgroup$
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
4 hours ago
add a comment |
8
8
8
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
answered 4 hours ago
saulspatzsaulspatz
17k31434
$endgroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
answered 4 hours ago
saulspatzsaulspatz
17k31434
answered 4 hours ago
saulspatzsaulspatz
17k31434
answered 4 hours ago
saulspatzsaulspatz
17k31434
answered 4 hours ago
saulspatzsaulspatz
17k31434
17k31434
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
4 hours ago
add a comment |
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
4 hours ago
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Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
5 hours ago
2
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
5 hours ago
1
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
5 hours ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
5 hours ago